# What is the vertex of  y= 5x^2-x-1+(2x-1)^2?

Dec 16, 2015

vertex$= \left(\frac{5}{18} , - \frac{25}{36}\right)$

#### Explanation:

Start by expanding the brackets and simplifying the expression.

$y = 5 {x}^{2} - x - 1 + {\left(2 x - 1\right)}^{2}$
$y = 5 {x}^{2} - x - 1 + \left(4 {x}^{2} - 4 x + 1\right)$
$y = 9 {x}^{2} - 5 x$

Take your simplified equation and complete the square.

$y = 9 {x}^{2} - 5 x$
$y = 9 \left({x}^{2} - \frac{5}{9} x + {\left(\frac{\frac{5}{9}}{2}\right)}^{2} - {\left(\frac{\frac{5}{9}}{2}\right)}^{2}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + {\left(\frac{5}{18}\right)}^{2} - {\left(\frac{5}{18}\right)}^{2}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + \frac{25}{324} - \frac{25}{324}\right)$
$y = 9 \left({x}^{2} - \frac{5}{9} x + \frac{25}{324}\right) - \left(\frac{25}{324} \cdot 9\right)$
$y = 9 {\left(x - \frac{5}{18}\right)}^{2} - \left(\frac{25}{\textcolor{red}{\cancel{\textcolor{b l a c k}{324}}}} ^ 36 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}\right)$
$y = 9 {\left(x - \frac{5}{18}\right)}^{2} - \frac{25}{36}$

Recall that the general equation of a quadratic equation written in vertex form is:

$y = a {\left(x - h\right)}^{2} + k$

where:
$h =$x-coordinate of the vertex
$k =$y-coordinate of the vertex

So in this case, the vertex is $\left(\frac{5}{18} , - \frac{25}{36}\right)$.